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In the preprint, A note presenting the generalization of a special logarithmic integral by Cornel Ioan Valean, it is given the following generalization,

Let $n\ge1$ be a positive integer. Then, \begin{equation*} \int_0^1 \frac{\log ^{2n-1}(x) \log(1-x)}{1+x} \textrm{d}x \end{equation*} \begin{equation*} =\frac{1}{2}(2n)!\zeta (2n+1)-2\log(2)(1 -2^{-2n})(2n-1)!\zeta (2n) \end{equation*} \begin{equation*} -2^{-1-2n} (2n+1-2^{1+2n})(2n-1)!\zeta(2n+1) \end{equation*} \begin{equation*} -(2n-1)!\sum_{k=1}^{n-1}\zeta (2k)\zeta (2n-2k+1)+2^{-2n}(2n-1)!\sum_{k=1}^{n-1}2^{2k}\zeta (2k)\zeta (2n-2k+1), \end{equation*} where $\zeta$ represents the Riemann zeta function.

Question 1: Is the present generalization known in the literature? I would kindly appreciate any source with precise information.

Interestingly, on page $4$ it is presented the case $n=1$ which surprisingly can be evaluated without using Beta function, Polylogarithm, or Euler sums.

Question 2: Is it possible to proceed in the same style for other cases of the generalization or for the whole generalization and get an elegant solution without using Beta function, Polylogarithm, or Euler sums?

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  • $\begingroup$ @DavidH Thank you! That would be very good to know. $\endgroup$ Sep 18, 2019 at 10:17

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Alternatively, the closed form may put in the form,

$$\int_0^1 \frac{\log ^{2n-1}(x) \log(1-x)}{1+x} \textrm{d}x$$ $$=\frac{1}{2^{2n+1}}((n+1)2^{2n+1}-2n-1)(2n-1)! \zeta(2n+1)-\log(2) (2n-1)!\zeta(2n )$$

$$-\left(1-\frac{1}{2^{2n-1}}\right)(2n-1)!\sum_{k=1}^{n-1} \zeta(2k)\zeta(2n-2k+1)$$ $$+(2n-1)!\sum_{k=1}^{n-1}\eta(2k)\zeta(2n-2k+1)-(2n-1)!\sum_{k=0}^{n-1} \eta (2k+1)\eta (2n-2k),$$ where $\zeta$ represents the Riemann zeta function and $\eta$ denotes the Dirichlet eta function.

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